Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

Analysis: Note that AbstractList and Integer classes have overridden the equals() method, so in the code below, if(!res.contians(clone)) res.add(clone); can be used to de-dup. 

The algorithm for this problem is almost the same as the Combination Sum problem, except equals test mentioned above. 

public class Solution {
	public void combinationSum2(int[] num, int index, int target, ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tem) {
		if(target == 0) {
			ArrayList<Integer> clone = new ArrayList<Integer>(tem);
			if(!res.contains(clone)) res.add(clone);
			return;
		}

		int i = index;
		while(i<num.length && target-num[i]>=0) {
			tem.add(num[i]);
			combinationSum2(num, i+1, target-num[i], res, tem);
			tem.remove(tem.size()-1);
			i++;
		}
		return;
	}

	public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
		ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
		ArrayList<Integer> tem = new ArrayList<Integer>();
		Arrays.sort(num);
		combinationSum2(num, 0, target, res, tem);
		return res;
    }
}